How to find out length of bisector?

Question

I found this challenge online but can't solve it.

I have a triangle C is 90 degrees. The bisector of A cuts BC into 5 and 3, how do i find out how long the bisector is?

Answer 1

Suppose $AD$ bisects $\angle BAC$. From the angle bisector theorem, we have $$\frac{BD}{DC} = \frac{AB}{AC} = \frac{5}{3}$$

Letting $l(AB) = 5x$ and $l(AC) = 3x$, by Pythagoras thoerem, $$25x^2 = 64 + 9x^2$$ $$\implies 16x^2 = 64$$ $$\implies x = 2$$

Therefore $l(AC) = 6$, $l(DC) = 3$, now we use Pythagoras theorem again to compute $l(AD)$, $$l(AD) = \sqrt{36+9} = \sqrt{45} = 3\sqrt{5}$$

Answer 2

From the given, $$\frac{\tan2\alpha}{\tan\alpha}=\frac{5+3}3,$$

and we are looking for the value of

$$\frac3{\sin\alpha}.$$

$$\tan\alpha=\frac12\to\sin\alpha=\frac1{\sqrt5},\\3\sqrt5.$$